GATE Papers >> CSE >> 2019 >> Question No 40

Question No. 40 CSE | GATE 2019

Consider three 4-variable functions f1, f2, and f3, which are expressed in sum-of-minterms as f1 = $\sum$ (0, 2, 5, 8, 14),   f2 = $\sum$ (2, 3, 6, 8, 14, 15),   f3 = $\sum$ (2, 7, 11, 14) For the following circuit with one AND gate and one XOR gate, the output function f can be expressed as:

 


Answer : (A) $\sum$ (7, 8, 11)


Solution of Question No 40 of GATE 2019 CSE Paper

f1 = m0 + m2 + m5 + m8 + m_14

f2 = m2 + m3 + m6 + m8 + m14 + m15

f3 = m2 + m7 + m11 + m14

⇒ f1 . f2 = m2 + m8 + m14

$ \mathrm f=\left({\mathrm f}_1.{\mathrm f}_2\right)\oplus{\mathrm f}_3 $

  A B C D f1 . f2 f3 F
m0 0 0 0 0 0 0 0
m1 0 0 0 1 0 0 0
m2 0 0 1 0 1 1 0
m3 0 0 1 1 0 0 0
m4 0 1 0 0 0 0 0
m5 0 1 0 1 0 0 0
m6 0 1 1 0 0 0 0
m7 0 1 1 1 0 1 1
m8 1 1 0 0 1 0 1
m9 1 1 0 1 0 0 0
m10 1 1 1 0 0 0 0
m11 1 1 1 1 0 1 1
m12 1 1 0 0 0 0 0
m13 1 1 0 1 0 0 0
m14 1 1 1 0 1 1 0
m15 1 1 1 1 0 0 0

$ \begin{array}{l}\Rightarrow\mathrm f={\mathrm m}_7+{\mathrm m}_8+{\mathrm m}_{11}\\\sum\mathrm m\left(7,8,11\right)\end{array} $

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